In one sense, there are only Kähler examples, but, in another sense, there are non-Kähler examples of such $J$. Here is what I mean:

Suppose that one has the data $(M,g,J)$ as defined in the question and that they satisfy the Conditions $1$ and $2$ (we'll get to $3$ below). Let $TM\otimes\mathbb{C} = W_+\oplus W_-$, where $W_+\subset TM\otimes\mathbb{C}$ is the $(+i)$-eigenspace of $J$ and $W_-\subset TM\otimes\mathbb{C}$ is the $(-i)$-eigenspace of $J$. (This direct sum decomposition exists because of Condition $1$.)

Now, by Condition $2$, $W_+$ and $W_-$ must be $g$-isotropic subbundles of $TM\otimes\mathbb{C}$. Since they have trivial intersection and sum to $TM\otimes\mathbb{C}$ (on which $g$ is nondegenerate), it follows that they must each be of half the dimension of $TM\otimes\mathbb{C}$, so, in particular, $M$ must have even dimension. Moreover, each of these spaces has trivial intersection with $TM\otimes 1$ and $TM\otimes i$ (since these latter spaces have no nonzero $g$-null vectors), so it follows that there are maps $K_\pm:TM\to TM$ such that
$$
W_\pm = \{\ v - i\ K_\pm v\ \bigl|\ v\in TM\ \}.
$$
Now, the condition that $W_\pm$ be $g$-null is equivalent to the condition that each of $K_+$ and $K_-$ be a $g$-orthogonal almost complex structure on $M$. The condition that these two spaces be disjoint is just that $K_+-K_-:TM\to TM$ have no kernel. Conversely, if $K_\pm:TM\to TM$ are two $g$-orthogonal almost complex structures on $M$ such that $K_+-K_-$ has no kernel, then defining the complex subbundles $W_\pm\subset TM\otimes\mathbb{C}$ as above and defining $J:TM\otimes\mathbb{C}\to TM\otimes \mathbb{C}$ to be the complex linear map that has $W_+$ as the $(+i)$-eigenspace of $J$ and $W_-$ as the $(-i)$-eigenspace of $J$ will give a triple $(M,g,J)$ that satisfies Conditions $1$ and $2$. Note that the $J$ that is so-defined comes from the almost complex structure $K_+$ in the usual way if and only if $K_- = - K_+$. Note that, in general, there could be many pairs $(K_+,K_-)$ that satisfy these conditions without satisfying $K_-=-K_+$.

Now, what about Condition 3 (i.e., $\nabla J=0$)? This is equivalent to $\nabla K_\pm = 0$, so one can get examples $(M, g, J)$ that satisfy the three Conditions without being derived from a standard Kähler structure only if $g$ has a pair of parallel complex structures $(K_+,K_-)$ that satisfy the above conditions but don't satisfy $K_-=-K_+$. This does happen; for example, if $g$ has holonomy $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$, then there is a $2$-sphere of $g$-parallel complex structures on $M$, and choosing any pair $(K_+,K_-)$ from that $2$-sphere that is not antipodal will allow one to generate a $(M,g,J)$ that does not come from a Kähler structure in the usual sense. However, note that the underlying Riemannian manifold *does* admit a Kähler structure, in fact, a $2$-sphere of them. The family of $(M,g,J)$ that one gets for such a manifold has dimension $4$ and contains the 'usual' Kähler structures as an embedded $2$-sphere. There can be a higher dimensional family of solution pairs if the holonomy of $g$ is a proper subset of $\mathrm{Sp}(n)$, but, in such cases, the metric will be a product (locally), and the problem reduces to working on the irreducible factors.